Problem with binary tree program

In my program, I pick a pivot p from a database S of strings, compute the median r of the distances of other string objects to p and then divide the other objects into roughly equal-sized subsets S1 and S2 as follows:

S1={o e S \{p}|d(p,o)<r}
S2={o e S \{p}|d(p,o)>=r}

where d(p,o) is the distance of the database object o to pivot p.Thus, the objects in S1 are inside the ball of radius r around p, while the objects in S2 are outside this ball. Applying this rule recursively leads to a binary tree,where a pivot object is stored in each internal node, with the left and right subtrees corresponding to the subsets inside and outside the corresponding ball,respectively.

Below I give the function for the above criteria but when I call the function from main and try to print the tree in preorder, nothing gets printed nor does the program terminate. I will really appreciate if someone can help me find the errors in my code and where I am going wrong:

Code:

void query_ball()
{
        
string str;
int i=0,p;
int arr[1000];
char B[40];
char C[40];
vector<string> myStrings(1000);
int r;
int n1;

ifstream file("text.txt"); //this file is the database containing 1000 strings
                    
for(p=0;p<1000;p++)
{
   getline(file,str);
   strcpy(B,str.c_str());
   myStrings[p]=B;   // copying the file strings to vector myStrings
}
srand(time(0));
int pos;
pos=rand()%1000;
strcpy(C,myStrings[pos].c_str());   // C is the pivot picked randomly

for(i=pos;i<=998;i++)                  // deleting element at myStrings[pos] which contains pivot C
myStrings[i]=myStrings[i+1];

myStrings[i]="";

        
for(p=0;p<999;p++)
{        
strcpy(B,myStrings[p].c_str());
arr[p]=edit(B,C);   // arr contains edit distances between other string objects and pivot C                          
}

/*edit function relates to edit distances of pivot from other string objects. An edit distance between 2 strings is found by changing one string to another with the minimum number of insertions, deletions and/or replacement of characters in the strings.*/

n1=(999+1)/2;  
r=arr[n1];   // r is the median of the distances of other objects to pivot C

for (p=-1; p<999; p++) //p starts with -1 to check first if Root is Null
{
                    
   if (Root == NULL)
{
   Root = new TreeNode();
   Root->val = C;
}
                 
else
{
         TreeNode* temp = Root;
         while(temp!=NULL)
         { strcpy(B,myStrings[p].c_str());
           

            if (edit(B,C)<r)               // checking distances inside ball of radius r
            {
               if (temp->LChild==NULL)
                  { TreeNode* t = new TreeNode();
                     t->val = B;
  
                     temp->LChild=t;  
            }
         else
               {
                              TreeNode* t = new TreeNode();
                              t->val = B;
                              temp->LChild = temp->LChild->t;
                 
               }
            }

            if (edit(B,C)>=r) // checking distances outside ball of radius r
            {

               if (temp->RChild==NULL)
                     {   TreeNode* t = new TreeNode();
                           t->val = B;
  
                        temp->RChild=t;
                     }

            else
               {
                     TreeNode* t = new TreeNode();
                     t->val = B;
                     temp->RChild = temp->RChild->t;
                             
               }
            }  
   }

}
}

}